Lue Vang Mr. Kellogg AP Physics - Pd. 7 18 April 2012 Performed On: 11-12 April 2012 Lab Partners: Brent Keath, Kevin Trinh, Lue Vang
Rotational Inertia Demonstrator
Purpose: This lab will study ways of finding angular acceleration, moment of inertia, and kinetic energy through theoretical calculations and through experimental methods.
Background: The rotation inertia demonstrator will spin as weight is attached and released from one of its three pulleys. The falling mass will have a constant acceleration, ensuring a constant force and torque. This being the case, angular acceleration and the likes can be found using rotational kinematic equations or the equation for angular acceleration. In addition, an unbalanced torque from unbalanced forces will cause an angular acceleration. Torque produced by tension on the string of the hanging mass can then be used to calculate moment of inertia by using Newton’s second law or by computing mass and distances from fix axes. Finally, kinetic energy is made of up both translational and rotational kinetic energy, so according to the law of conservation of energy, the potential energy before dropping the mass can be used to determine how much of the total energy is converted to translational kinetic energy and how much is converted to rotational energy.
Hypothesis: The two methods of calculating angular acceleration should produce similar, if not equal, answers. Likewise, the two methods for calculation the moment of inertia for the pulley should equate to similar, if not equal, answers. Finally, for the calculations of energy, the total energy from translational kinetic energy and rotational energy should equal to original potential energy.
Apparatus:
Rotational inertia demonstrator (1)
Stop watch (1)
Mass set (1)
String (1m)
Meter stick (1)
Scale (1)
Calipers (1)
Procedure: Experiment 1: Kinematics of Rotational Motion Hang a mass from one end of the string and loop the string around one of the rotational inertia demonstrator (RID)’s, three pulleys. Let the mass drop to the floor, and wind it up with 4-6 rotations. Measure the vertical distance of the drop and the time required for the mass to fall so. Use the data to calculate rotational acceleration. Experiment 2: Determining the Moment of Inertia of the Rotational Inertia Demonstrator Find the linear and rotational acceleration of the falling mass. Then, calculate the torque produced by tension. Once found, use the said data to calculate moment of inertia (I). Afterwards, find the total moment of inertia by computing and adding the moment of inertia of all the bodies in the pulley system. Finally, compare the two calculated moments of inertia. Experiment 3: Energy of Rotational Motion Measure the mass and height of the hanging mass at starting position to calculate the total potential energy. Next, time the mass’s descent to the floor and use it to calculate the velocity, which will then we used to calculate translational kinetic energy. Similarly, use the angular velocity to find the total rotational kinetic energy of the system.
Data: Experiment 1: Kinematics of Rotational Motion
a = 2(Δs – v0t)/t2 a = 2(0.92 – 0*9.5)/ 9.52 a ≈ 0.0204 m/s2 Pulley Radius: 0.03 m
α = a/r⊥ α = 0.0204/0.03 α ≈ 0.68 rad/s2
Percent difference (α) = |(difference of α/average α)|*100 = |(.68-.6962)/((.68+.6962)/2)|*100 ≈ 2.35% = Close enough
Experiment 2: Determining the Moment of Inertia of the Rotational Inertia Demonstrator Moment of Inertia: Method 1 Angular acceleration of rotating apparatus: 0.70 rad/s2 Linear acceleration of falling mass: 0.02 m/s2
Radius of pulley: 0.03 m Mass of falling mass: 0.23 kg
Tension = -(Weight) = -(mg) = -(0.23*-9.8) = 2.254 N
Moment of Inertia: Method 2 Pulleys: I = 0.00058 kg*m2 Thin Rods: Mass: 0.0748 kg Length: 0.345 m I per rod = 1/3ML2 = (1/3)0.0748*0.345^2 ≈ 0.003 kg*m2 I for four rods = 4(1/3ML2) = 4((1/3)0.0748*0.345^2) ≈ 0.012 kg*m2 Movable Mass: Mass: 0.018 kg Distance from Center (radius): 0.02125 m I per movable mass = 1/2MR2 =1/2(0.018*0.021252) ≈ 0.009*0.021252 kg*m2 I for four movable masses = 4(1/2MR2) = 4(1/2(0.018*0.021252)) ≈ 0.0000162 kg*m2 ΣI = I (pulley) + I (rods) + I (movable masses) = 0.00058 + 0.012 + .000016 ≈ 0.013 kg*m2
Experiment 3: Energy of Rotational Motion Falling Mass: Mass: .23 kg Vertical displacement: .9m Potential energy = mgh = .23*9.8*.9 = 2.0286 J Fall time: 9.53 s Average velocity = d/t = 0.9 /9.53 ≈ 0.094 m/s Final velocity = 2vavg = 2(0.094) ≈ 0.189 m/s
Analysis: In experiment 1, the angular acceleration obtained from kinematic equations is about 0.70 rad/s2 while the one from using (α = a/r) is about 0.68 rad/s2. The percent difference between the two is 2.35%, so the two results are not significantly different since a small change as +0.03 rad/s2 to 0.68 rad/s2 would result in a large of a decrease as about 5%. In fact, that means the results are very close, so the data suggests that methods 1 and 2 agree within the uncertainty of measurement.
In experiment 2, the two calculated moments of inertia should be of close values, and they are. The calculated method resulted in a moment of inertia of 0.971 kg*m2 while the experimental method produced a moment of inertia of 0.013 kg*m2. The percent difference between the two methods is 152.77%, so according to the lab results, it is method 1 and 2 do not agree within the uncertainty of measurement.
The calculative method resulted in a total energy of about 2.03 J while the experimental method produced a value of about 1.93 J. In comparison, the two are only different by a percentage of about 4.77%, thus methods 1 and 2 produce values that agree within the uncertainty of measurement.
Conclusion: All in all, the lab appears to be a relative success. Ideally, the calculations should end up with close results, but experiment 2’s did not.
Experiment 1 was pretty good with only a 2.35% difference. The only plausible source of inaccuracy may be from the measurements taken. Also, friction may skew the data a tad. Still, the percent difference assures that the two values from the two different methods agree within the uncertainty of measurement.
Experiment 2 resulted in a unexpectedly high percent difference. This may be due to a inadequate collection of data because since the equation for the moment of inertia of disks is (1/2)MR2, the radius of the disks were collected for “distance from the center” in method 2. It is unclear whether the lab wanted the radius of the disk or the distance of the center of mass of the disk from the center of the the system. Nonetheless, substituting a reasonable value of .02125m – 0.355m would’ve still ended up with a moment of inertia with a large percentage difference from method 1’s moment of inertia. The percent difference would’ve still been over 100%.
Experiment 3 ended up with a good percent difference of only 4.77%, is very accurate. Again, like experiment 1, the difference may be a result of slightly inaccurate measures and/or the loss of energy through friction. Still, the low percent difference indicates that the results agree within the uncertainty of measurement.
Seeing as the calculations of the proceeding experiment depends on the calculations of the experiment preceding it, miscalculation from one section of this lab will greatly result in miscalculations in the experiment after.
Overall, experiment 1and 3 turned out relatively well, but experiments 2 was not successful. To increase the chances of success in the future, it would be wise to take better notes when collecting data or be more specific in the lab directions. Pictures, diagrams, and/or video documentation would be most useful as well.
Mr. Kellogg
AP Physics - Pd. 7
18 April 2012
Performed On: 11-12 April 2012
Lab Partners: Brent Keath, Kevin Trinh, Lue Vang
Rotational Inertia Demonstrator
Purpose:
This lab will study ways of finding angular acceleration, moment of inertia, and kinetic energy through theoretical calculations and through experimental methods.
Background:
The rotation inertia demonstrator will spin as weight is attached and released from one of its three pulleys. The falling mass will have a constant acceleration, ensuring a constant force and torque. This being the case, angular acceleration and the likes can be found using rotational kinematic equations or the equation for angular acceleration.
In addition, an unbalanced torque from unbalanced forces will cause an angular acceleration. Torque produced by tension on the string of the hanging mass can then be used to calculate moment of inertia by using Newton’s second law or by computing mass and distances from fix axes.
Finally, kinetic energy is made of up both translational and rotational kinetic energy, so according to the law of conservation of energy, the potential energy before dropping the mass can be used to determine how much of the total energy is converted to translational kinetic energy and how much is converted to rotational energy.
Hypothesis:
The two methods of calculating angular acceleration should produce similar, if not equal, answers.
Likewise, the two methods for calculation the moment of inertia for the pulley should equate to similar, if not equal, answers.
Finally, for the calculations of energy, the total energy from translational kinetic energy and rotational energy should equal to original potential energy.
Apparatus:
Procedure:
Experiment 1: Kinematics of Rotational Motion
Hang a mass from one end of the string and loop the string around one of the rotational inertia demonstrator (RID)’s, three pulleys. Let the mass drop to the floor, and wind it up with 4-6 rotations. Measure the vertical distance of the drop and the time required for the mass to fall so. Use the data to calculate rotational acceleration.
Experiment 2: Determining the Moment of Inertia of the Rotational Inertia Demonstrator
Find the linear and rotational acceleration of the falling mass. Then, calculate the torque produced by tension. Once found, use the said data to calculate moment of inertia (I). Afterwards, find the total moment of inertia by computing and adding the moment of inertia of all the bodies in the pulley system. Finally, compare the two calculated moments of inertia.
Experiment 3: Energy of Rotational Motion
Measure the mass and height of the hanging mass at starting position to calculate the total potential energy. Next, time the mass’s descent to the floor and use it to calculate the velocity, which will then we used to calculate translational kinetic energy. Similarly, use the angular velocity to find the total rotational kinetic energy of the system.
Data:
Experiment 1: Kinematics of Rotational Motion
Angular Acceleration: Method 1
α = 2(ΔΘ – ω0t)/t2
α = 2(10π – 0*9.5)/ 9.52
α ≈ 0.6962 rad/s2
Angular Acceleration: Method 2
a = αr
Vertical Displacement: 0.92m
a = 2(Δs – v0t)/t2
a = 2(0.92 – 0*9.5)/ 9.52
a ≈ 0.0204 m/s2
Pulley Radius: 0.03 m
α = a/r⊥
α = 0.0204/0.03
α ≈ 0.68 rad/s2
Percent difference (α) = |(difference of α/average α)|*100
= |(.68-.6962)/((.68+.6962)/2)|*100
≈ 2.35% = Close enough
Experiment 2: Determining the Moment of Inertia of the Rotational Inertia Demonstrator
Moment of Inertia: Method 1
Angular acceleration of rotating apparatus: 0.70 rad/s2
Linear acceleration of falling mass: 0.02 m/s2
Radius of pulley: 0.03 m
Mass of falling mass: 0.23 kg
Tension = -(Weight) = -(mg)
= -(0.23*-9.8)
= 2.254 N
τ = Iα
I = τ/α = (r⊥T)/α
= (0.03*2.254)/0.6962
= 0.0971 kg*m2
Moment of Inertia: Method 2
Pulleys:
I = 0.00058 kg*m2
Thin Rods:
Mass: 0.0748 kg
Length: 0.345 m
I per rod = 1/3ML2
= (1/3)0.0748*0.345^2
≈ 0.003 kg*m2
I for four rods = 4(1/3ML2)
= 4((1/3)0.0748*0.345^2)
≈ 0.012 kg*m2
Movable Mass:
Mass: 0.018 kg
Distance from Center (radius): 0.02125 m
I per movable mass = 1/2MR2
=1/2(0.018*0.021252)
≈ 0.009*0.021252 kg*m2
I for four movable masses = 4(1/2MR2)
= 4(1/2(0.018*0.021252))
≈ 0.0000162 kg*m2
ΣI = I (pulley) + I (rods) + I (movable masses)
= 0.00058 + 0.012 + .000016
≈ 0.013 kg*m2
Percent difference (α) = |(difference of α/average α)|*100
= |(.0971-.013)/((.0971+.013)/2)|*100
≈ 152.77%
Experiment 3: Energy of Rotational Motion
Falling Mass:
Mass: .23 kg
Vertical displacement: .9m
Potential energy = mgh
= .23*9.8*.9
= 2.0286 J
Fall time: 9.53 s
Average velocity = d/t
= 0.9 /9.53
≈ 0.094 m/s
Final velocity = 2vavg
= 2(0.094)
≈ 0.189 m/s
Translational Kinetic Energy:
KEf = (1/2)mv2
= (1/2)(.23*0.1892)
=.0041 J
Pulley:
Radius: 0.03 m
ωf = v/r
= 0.189/0.03
= 6.3 rad/s
I = 0.0971
Rotational Kinetic Energy:
KE = (1/2)Iω2
= (1/2)0.0971*6.32
= 1.93 J
KE (total) = KEtrans + KErot
= 0.0041 + 1.93
= 1.9341 J
Percent difference (E) = |(difference of E/average E)|*100
= |((2.0286-1.9341)/((2.0286+1.9341)/2)|*100
≈ 4.77%
Analysis:
In experiment 1, the angular acceleration obtained from kinematic equations is about 0.70 rad/s2 while the one from using (α = a/r) is about 0.68 rad/s2. The percent difference between the two is 2.35%, so the two results are not significantly different since a small change as +0.03 rad/s2 to 0.68 rad/s2 would result in a large of a decrease as about 5%. In fact, that means the results are very close, so the data suggests that methods 1 and 2 agree within the uncertainty of measurement.
In experiment 2, the two calculated moments of inertia should be of close values, and they are. The calculated method resulted in a moment of inertia of 0.971 kg*m2 while the experimental method produced a moment of inertia of 0.013 kg*m2. The percent difference between the two methods is 152.77%, so according to the lab results, it is method 1 and 2 do not agree within the uncertainty of measurement.
The calculative method resulted in a total energy of about 2.03 J while the experimental method produced a value of about 1.93 J. In comparison, the two are only different by a percentage of about 4.77%, thus methods 1 and 2 produce values that agree within the uncertainty of measurement.
Conclusion:
All in all, the lab appears to be a relative success. Ideally, the calculations should end up with close results, but experiment 2’s did not.
Experiment 1 was pretty good with only a 2.35% difference. The only plausible source of inaccuracy may be from the measurements taken. Also, friction may skew the data a tad. Still, the percent difference assures that the two values from the two different methods agree within the uncertainty of measurement.
Experiment 2 resulted in a unexpectedly high percent difference. This may be due to a inadequate collection of data because since the equation for the moment of inertia of disks is (1/2)MR2, the radius of the disks were collected for “distance from the center” in method 2. It is unclear whether the lab wanted the radius of the disk or the distance of the center of mass of the disk from the center of the the system. Nonetheless, substituting a reasonable value of .02125m – 0.355m would’ve still ended up with a moment of inertia with a large percentage difference from method 1’s moment of inertia. The percent difference would’ve still been over 100%.
Experiment 3 ended up with a good percent difference of only 4.77%, is very accurate. Again, like experiment 1, the difference may be a result of slightly inaccurate measures and/or the loss of energy through friction. Still, the low percent difference indicates that the results agree within the uncertainty of measurement.
Seeing as the calculations of the proceeding experiment depends on the calculations of the experiment preceding it, miscalculation from one section of this lab will greatly result in miscalculations in the experiment after.
Overall, experiment 1and 3 turned out relatively well, but experiments 2 was not successful. To increase the chances of success in the future, it would be wise to take better notes when collecting data or be more specific in the lab directions. Pictures, diagrams, and/or video documentation would be most useful as well.